RD Sharma Class 7 ex 16.2 Solutions Chapter 16 Congruence

In this chapter, we provide RD Sharma Class 7 ex 16.2 Solutions Chapter 16 Congruence for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 16.2 Solutions Chapter 16 Congruence Maths pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 16
Chapter NameProperties of Triangles
ExerciseEx 16.2

RD Sharma Solutions for Class 7 Chapter 16 Congruence Ex 16.2 Download PDF

Congruence Exercise 16.2

Question: 1

In the following pairs of triangle (Figures), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.

Congruence Exercise 16.2 Question: 1A
Congruence Exercise 16.2 Question: 1B

Solution:

Congruence Exercise 16.2 Solution: 1A

1) In Δ ABC and Δ DEF

AB = DE = 4.5 cm (Side)

BC = EF = 6 cm (Side) and

AC = DF = 4 cm (Side)

Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF

2)

Congruence Exercise 16.2 Solution: 1B

In Δ ACB and Δ ADB

AC = AD (Side)

BC = BD (Side) and

AB = AB (Side)

Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB

3) In Δ ABD and Δ FEC,

AB = FE (Side)

AD = FC (Side)

BD = CE (Side)

Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC

4) In Δ ABO and Δ DOC,

AB = DC (Side)

AO = OC (Side)

BO = OD (Side)

Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC

Question: 2

In figure, AD = DC and AB = BC

(i) Is ΔABD ≅ ΔCBD?

(ii) State the three parts of matching pairs you have used to answer (i).

Congruence Exercise 16.2 Question: 2

Solution:

Yes ΔABD = ΔCBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:

AD = DC

AB = BC and

DB = BD

Question: 3

In Figure, AB = DC and BC = AD.

(i) Is ΔABC ≅ ΔCDA?

(ii) What congruence condition have you used?

(iii) You have used some fact, not given in the question, what is that?

Congruence Exercise 16.2 Question: 3

Solution:

We have AB = DC

BC = AD

and AC = AC

Therefore by SSS ΔABC ≅ ΔCDA

We have used Side congruence condition with one side common in both the triangles.

Yes, have used the fact that AC = CA.
 

Question: 4

In ΔPQR ≅ ΔEFD,

(i) Which side of ΔPQR equals ED?

(ii) Which angle of ΔPQR equals angle E?

Solution:

ΔPQR ≅ ΔEFD

(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.

(ii) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.

Congruence Exercise 16.2 Question: 4


Question: 5

Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?

It ∠B = 50°, what is the measure of ∠R?

Solution:

We have AB = AC in isosceles ΔABC

And PQ = PR in isosceles ΔPQR.

Also, we are given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)

Hence, ΔABC ≅ ΔPQR

Now

∠ABC = ∠PQR (Since triangles are congruent)However, ΔPQR is isosceles.

Therefore, ∠PRQ = ∠PQR = ∠ABC = 50°

Question: 6

ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.

Solution:

∠BAD = ∠CAD (c.p.c.t)

∠BAD + ∠CAD = 40°/ 2 ∠BAD = 40°

∠BAD = 40°/2 =20°

∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)

Since ΔABC is an isosceles triangle,

∠ABC = ∠BCA ∠ABC +∠ABC + 40°= 180°

2 ∠ABC = 180° – 40° = 140° ∠ABC = 140°/2 = 70°

∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)

Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD ∠DBC + ∠DBC + 100= 180°

2 ∠DBC = 180° – 100o = 80°

∠DBC = 80°/2 = 40°

In Δ BAD,

∠ABD + ∠BAD + ∠ADB = 180°(Angle sum property)

30° + 20° + ∠ADB = 180° (∠ADB = ∠ABC – ∠DBC), ∠ADB = 180°- 20° – 30°

∠ADB = 130°

∠ADB =130°

Question: 7

Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Figure. Which of the following statements is true?

(i) ΔABC ≅ ΔABD

(ii) ΔABC ≅ ΔADB

(iii) ΔABC ≅ ΔBAD

Congruence Exercise 16.2 Question: 7

Solution:

In ΔABC and ΔBAD we have,

AC = BD (given)

BC = AD (given)

and AB = BA (common)

Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD

There option (iii) is true.

Question: 8

In Figure, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.

(i) Is ΔADB ≅ ΔADC?

(ii) State the three pairs of matching parts you use to arrive at your answer.

Congruence Exercise 16.2 Question: 8

Solution:

We have AB = AC.

Also since D is the midpoint of BC, BD = DC

Also, AD = DA

Therefore by SSS condition,

ΔADB ≅ ΔADC

We have used AB, AC : BD, DC AND AD, DA
 

Question: 9

In figure, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.

Congruence Exercise 16.2 Question: 9

Solution:

Yes, ΔABC ≅ ΔACB by SSS condition.

Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB

Question: 10

Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?

Solution:

Yes,

Given,

Δ ABC and Δ DBC have side BC common, AB = BD and AC = CD

By SSS criterion of congruency, ΔABC ≅ ΔDBC

No, ∠ABD and ∠ACD are not equal because AB ≠ AC

All Chapter RD Sharma Solutions For Class 7 Maths

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